Optics B/C

[JSGandora]

Oh, funny. I changed that purposely because after glancing at the problem statement, I thought I had the wrong numbers. In that case, Infinity Flat is correct, and this is the Wolfram Alpha link:

http://www.wolframalpha.com/input/?i=5s ... %29*10%5E6

[2win]

When the rules say you can't mark on or modify the LSS, does that count barriers?

Like, what if my template requires lifting a barrier to put the template underneath, but I would put the barrier back exactly in its place... is that allowed?

[Infinity Flat]

Oh, funny. I changed that purposely because after glancing at the problem statement, I thought I had the wrong numbers. In that case, Infinity Flat is correct, and this is the Wolfram Alpha link:

http://www.wolframalpha.com/input/?i=5s ... %29*10%5E6

Alright, great.
While we're on topic, was everything else on the test OK? Aside from being a bit on the difficult side.

When the rules say you can't mark on or modify the LSS, does that count barriers?

Like, what if my template requires lifting a barrier to put the template underneath, but I would put the barrier back exactly in its place... is that allowed?

You cannot touch the barriers.
Instead, you need to design your template in such a way that it goes around the barriers. Alternatively, you can cut a hole in it at the competition for the barriers to go through.

[JSGandora]

Incidentally, the Camas test that I took (a couple months back) looked nothing like the link it is now. So...I'll take that soon and I'll get back to you.

[cngu23]

Incidentally, the Camas test that I took (a couple months back) looked nothing like the link it is now. So...I'll take that soon and I'll get back to you.

The test was for division B.

For the question on speed of light, I think that the answer should be zero because light cannot pass through certain solid materials, so it would have a speed of zero?

And for the last question about total internal reflection, what are the requirements for that to occur. Just sin(theta) equals 1?
1.75sin(59) = 1.52sin(theta)
For total internal reflection to occur, do both sides have to be equal, with one side being theta = 90?

[foreverphysics]

Well, the formula for the critical angle is this:

So...solve with that, I suppose. I don't know what the last question was, so all I can do is give you a formula.

[JSGandora]

Well first use that formula to find the critical angle. Then all incidence angles greater than that will totally internally reflect (if I may).

[Seracon]

Another questions about Thin Films;
A silicon monoxide thin film (n=1.45) with thickness 90nm is applied to a camera lens made of glass (n=1.55). This will result in destructive interference for reflected light of what wavelength?
Thanks again!

[cngu23]

Another questions about Thin Films;
A silicon monoxide thin film (n=1.45) with thickness 90nm is applied to a camera lens made of glass (n=1.55). This will result in destructive interference for reflected light of what wavelength?
Thanks again!

31.03 nm? I'm not very sure, but just for the silicon monoxide, destructive interference occurs at even intervals so it should occur at Lambda/2n

[Seracon]

Another questions about Thin Films;
A silicon monoxide thin film (n=1.45) with thickness 90nm is applied to a camera lens made of glass (n=1.55). This will result in destructive interference for reflected light of what wavelength?
Thanks again!

31.03 nm? I'm not very sure, but just for the silicon monoxide, destructive interference occurs at even intervals so it should occur at Lambda/2n

Yeah, I'm not sure if having the index of refraction for the camera lens affects anything or not.